Splitting/Exploding object & Momentum

[Deadawake]

Hi,

Homework Statement

A free falling object of mass "m" falling from some height, collides the floor in speed of 20 m/s (perfectly elastic collision). In his 1/2 height back up he splits into 2 pieces- ¼m which going downward and ¾m keeping upward. The ¼m reaching the floor after ½ second.
1) What is the object velocity right before the split?
2) What is the velocity of the small object (¼m) right after the split?
3) What is the velocity of the big object (¾m) right after the split?

Homework Equations

The Attempt at a Solution

So I used conservation of energy to solve the first question. the initial height is 20.4m. it means the splitting point happened at 10.2m.
mgh = ½mv² , v= 14.14 m/s

Now I'm looking on the second question, because I know the falling time I can use kinematic equation.
0 = 10.2 +v₀t +½at²
-10.2 = 0.5v₀ + ½⋅(-9.8) ⋅0.5²
-7.55 = 0.5v₀
v₀ = -17.95

To figure out the big object velocity I used the coservation of momentum and here is where I got stuck

if a already have a negative velocity which determine the direction should I add a minus sign to the conservation of momentun equation too?
-¼mv₀ + ¾mV₀ = m⋅14.14
here when I put into the equation the negative velocity of v I get V₀ = 12.87

in the other hand , with positive momentum :
+¼mv₀ + ¾mV₀ = m⋅14.14

I get V₀ = 24.83

For some reason I think the first equation answer is more logical , It doesn't make sense that the ¾m accelarated too much after the split.
But it feels wrong when I get negative velocity and in the momentum equation I need to add negative sign as well .The negative velocity will do it anyway, won't it?

Thanks a lot!

 

[haruspex]

It doesn't make sense that the ¾m accelarated too much after the split.

But it makes even less sense that it moves more slowly after the split.

should I add a minus sign to the conservation of momentun equation too?

Certainly not. The conservation equation does not care whether the values to be plugged in are positive or negative, the equation remains the same.

 

[Deadawake]

But it makes even less sense that it moves more slowly after the split.

Certainly not. The conservation equation does not care whether the values to be plugged in are positive or negative, the equation remains the same.

Thanks. If it doesn't care about the values inside why it gives me different answers ?

 

[haruspex]

Thanks. If it doesn't care about the values inside why it gives me different answers ?

Because you changed the sign in the algebraic equation. The equation, as an algebraic statement, is the same whether the values are positive or negative.
If you have an equation x+y=z, and you are told x=-1, that does not change the equation to be -x+y=z.

 

[Deadawake]

Because you changed the sign in the algebraic equation. The equation, as an algebraic statement, is the same whether the values are positive or negative.
If you have an equation x+y=z, and you are told x=-1, that does not change the equation to be -x+y=z.

Thanks a lot.
Is it logical that after the explosion/splitting the bigger mass has more kinetic energy than the smaller mass? this is what I got here and it also doesn't make sense to me.

 

[haruspex]

Is it logical that after the explosion/splitting the bigger mass has more kinetic energy than the smaller mass?

Yes, that can happen. Consider e.g. if the explosion had been exactly enough to halt the smaller mass. It would have lost all its KE as a result, while the larger mass would have gained KE.